Posted June 22, 2022 by heaermo

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ESysBMWCodingv324364bit

ESysBMWCodingv324364bit

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ESysBMWCodingv324364bit
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ESysBMWCodingv324364bit.Q:

Inverse of the following function

I have this function $f(z)=z^n$ with $z\in \mathbb{C}$ and $n\in \mathbb{N}$.
I would like to get the inverse of $f$ for some specific values of $n$ and specific $z$.
My attempts:
I’ve tried using the function $f'(z)=nz^{n-1}$, but I’m getting complex numbers that don’t seem to be the right ones.

A:

Hint: There is no real $z$ such that $\mathrm{Re}\, z^n = 0$, and $\mathrm{Im}\, z^n = 0$. And there is no real $w$ such that $z = e^w$.

A:

The function is not invertible, because you can find $z$ with $z^n=\pm e^{i\theta}$ which are all in the same Riemann surface.
For $n=0$ the inverse is:
$$f(z)=z\iff z=e^{it},t\in\mathbb R.$$
For $n=1$ the inverse is:
$$f(z)=z\iff \mathrm{Re}z=0.$$
As for the inverses when $z=e^{i\theta}$ is a complex root, then there exists $\lambda\in\mathbb R$ such that $\lambda+i\theta=\lambda i$.
So,
$$\lambda+i\theta=i\cdot\lambda=\lambda i+i\theta=\lambda^2+\theta i=\lambda^2+i\theta,$$
which implies $\lambda=\pm i$ and thus
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